The beam A B supports two concentrated loads and rests on soil that exerts a linearly distribute

The beam A B supports two concentrated loads and rests on...

Key Concepts

Static Equilibrium

In mechanics, static equilibrium occurs when a body is at rest, meaning the vector sum of all forces and the sum of all moments about any point are zero. This principle is fundamental for analyzing structures and determining unknown reactions in any system under load.

Free Body Diagram (FBD)

A free body diagram is a simplified representation that isolates a structure or body and illustrates all the forces and moments acting on it. This visual tool is crucial for setting up the correct equilibrium equations needed to solve for unknown reactions.

Concentrated and Distributed Loads

Concentrated loads are forces acting at specific, discrete points, whereas distributed loads are spread continuously along a length of the structure. For analysis, distributed loads can often be replaced by an equivalent resultant force acting at a centroid, simplifying the process of establishing equilibrium.

Moment Equilibrium

Moment equilibrium involves taking moments about a chosen point to ensure that the turning effects of all forces cancel each other out. This is a key strategy for solving problems with multiple loads, including both point loads and distributed loads, allowing for the determination of unknown forces or reactions.

Transcript

00:04 In this problem, we have two discrete downward force vectors acting on a beam, under which we have a linearly increasing response density force along the bottom, the opposite side of the beam pointing in the opposite direction.

00:34 The distinct force vectors are 0 .6 meters to the left, i'm sorry, to the right of far left, and the second one is 30 kilonutons, 0 .3 meters from the far right.

01:02 The entire span of the beam is 1 .8 meters.

01:08 Here, given this linearly distributed upward force, we are tasked with determining the values of w sub a and w sub b.

01:19 These are force density values at the extreme ends of the upward force density relation.

01:33 We're tasked with finding those such that this system would be in equilibrium with the combined force of the two downward force vectors.

01:46 To solve this problem, we'll find the centroid of the two downward force vectors.

01:56 We'll then write the upward force density as a function of x and parameters w sub a and w sub b.

02:07 Third, we'll integrate to find the total load and total torque equation.

02:15 And these will be in terms of w sub a and w sub b.

02:22 Given both of those, we can then solve for these two unknowns, w sub a, w sub b.

02:29 Each of these will have units of kilonutons per meter.

02:36 They are basically the y coordinates of a force density relation that spans the length of the beam.

02:49 So for step one, we're finding the centroid and the total force from the two downward vectors.

02:57 To find the total load, we simply add the two, 24 plus 30 is 54 and we have units of kilonutons here.

03:07 To find the centroid of the two downward vectors, we multiply each of their forces by their respective moment arms as measured by the extreme left of the beam.

03:19 So we have 0 .6 meters times 24 kilonutons for the top, for the leftward vector torque.

03:30 We have 1 .5 meters times 30 kilonutons for the rightward vector torque.

03:37 We divide both of those, we divide the sum of those, sorry, in the numerator, by the denominator of 54 kiloons, which is the total force load magnitude and we're left with 1 .1 meters for the centroid of the beam from the downward forces.

03:57 In step two, we're going to write a relation for the linearly varying density of which points upward along the span of the beam.

04:12 Looking at our diagram here, our final y coordinate is w sub b.

04:20 Our initial y coordinate is w sub a.

04:23 And we're going to divide that difference between the two, w sub b minus w sub a, by the length that's separating them.

04:33 That is equal to the span of the beam and it's 1 .8 meters.

04:39 That quantity is our slope.

04:41 And for a linear relationship, we multiply.

04:45 By the slope the dependent variable x and that is the coordinate along the beam and we need to find an offset as well our offset is the y coordinate when x equals zero and for us that is simply w sub a so we now have our total we have the force density written as a linear function here as lowercase w of x times the difference divided by 1 .8, the difference between the two forced density values at the extreme ends of the beam plus the offset w sub a.

05:40 In step three, we're going to now derive equations for the total load and the centroid on the beam using the function.

05:56 We just derived the linear function of the force density that's pointing upward along the span of the beam...

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